Question

CMR :\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+......+\frac{1}{n^2}< 1\)

(n Thuoc N;n lon hon hoac = 2

Answer 1

Ta đặt:A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...\frac{1}{n^2}\)

Vì \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)

     \(\frac{1}{3^2}< \frac{1}{2\cdot3}\)

....

     \(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)

=> A < \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{\left(n-1\right)n}\)

=> A < \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)

=> A < \(1-\frac{1}{n}< 1\)(ĐPCM )

Vậy A < 1

Answer 2

Chững minh sao bạn !!!!!!!!!!!